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There is a useful result when studying lattices that the sum $\sideset{}{'}\sum_{m,n}(m^2 + n^2)^{-r/2}$ converges for $r > 2$, where the sum is taken over all pairs of integers $(m,n)\neq(0,0)$. I'm curious whether the condition $r > 2$ is really necessary or not. Does the sum $$\sideset{}{'}\sum_{m,n}\frac{1}{m^2 + n^2}$$ converge? Can we say what happens for fractional powers, or more generally, for which $x > 0$ the sum $$\sideset{}{'}\sum_{m,n}\frac{1}{(m^2 + n^2)^x}$$ converges?

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    $\begingroup$ One relevant fact is that every prime of the form $4k+1$ can be written as the sum of two squares. So a lower bound for the first sum is $\sum_{p\equiv1\pmod4} 1/p$, which diverges. $\endgroup$ Jul 21, 2023 at 22:04
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    $\begingroup$ just curious, what does the primed sum notation $\sideset{}{'}\sum_{m,n}$ mean? (finding it hard to search for this... searching for 'primed sum' gets me info on sums of primes) $\endgroup$
    – postylem
    Jul 22, 2023 at 13:37
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    $\begingroup$ @postylem The prime means to take the sum over all lattice points except for $(0,0)$. I've seen this notation used before a few times although I can't come up with a reference. $\endgroup$
    – Carmeister
    Jul 22, 2023 at 18:22
  • $\begingroup$ Ah okay! I think I've seen similar notation used for other cases where the sum is "not over the whole space" for some meaning relevant to the specific use case. But I also can't recall where I saw that. $\endgroup$
    – postylem
    Jul 23, 2023 at 2:19

5 Answers 5

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We can use Euler products to rewrite the sum as the product \begin{align} \sum_{\mathbb{Z}^2 \ni (m, n) \neq (0, 0)} \frac{1}{m^2+n^2} &= 4 \sum_{k \geq 1} \frac{r_2(k)}{4k} \\ &= 4 (1 + 2^{-1})^{-1} \prod_{p = 4j+1} (1 + p^{-1})^{-2}, \end{align} where $r_2(n)$ is the number of representations of $n$ as a sum of two squares and the last product is taken over primes. The second product diverges by Dirichlet's theorem on arithmetic progressions: $$ \prod_{p = 4j+1} (1 + p^{-1})^{-2} = \biggl[\,\prod_{p = 4j+1} (1 + p^{-1}) \biggr]^{-2} $$ diverges, because $$ \sum_{p = 4j+1} \frac{1}{p} $$ diverges.

So the set of all real numbers for which the sum $$ \sum_{\mathbb{Z}^2 \ni (m, n) \neq (0, 0)} (m^2+n^2)^{-r/2} $$ converges, really is the interval $r \in (2, \infty)$.

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For rough estimates you only need to know that the map

$$(x,y) \mapsto \sqrt{x^2 + y^2}$$ is a norm so comparable with any other norm on $\mathbb{R}^2$.

Hence you can ask for a ( any ) norm $\|\cdot \|$ on $\mathbb{R}^2$ for which $s$ the series

$$\sideset{}{'}\sum_{m,n} \frac{1}{\|(m,n)\|^s}$$

is convergent. Now choose a convenient norm, $\|\cdot \|_{\infty}$. The norm of every integral point is a natural number, and moreover each sphere of radius $N$ has an easy to see number of integral elements which is of order $N$. So now we have to see for which $s$ the series

$$\sum_{N\ge 1}\frac{N}{N^s}$$ which gets us $s > 2$.

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There are a variety of ways to show that $\sideset{}{'}\sum_{\!m,n} \frac{1}{m^2 + n^2}$ diverges depending on how the limit is taken. Here is one that shows that the sum over the first $n^2$ integer pairs from $1$ to $n$ grows like $\ln(n)$ and so diverges.

In particular, this shows that the sum over $1\le m\le N, 1\le n \le N $ grows like $\frac{\pi}{4}\ln(N)+O(1)$.

I have checked the estimate for $s_1(n)$, below, numerically.

\begin{align} s(n) &=\sum_{k=1}^n \sum_{j=1}^n \dfrac{1}{k^2 + j^2}\\ &=\sum_{k=1}^n \left(\sum_{j=1}^k \dfrac{1}{k^2 + j^2}+\sum_{j=k+1}^n \dfrac{1}{k^2 + j^2}\right)\\ &=\sum_{k=1}^n \sum_{j=1}^k \dfrac{1}{k^2 + j^2}+\sum_{k=1}^n \sum_{j=k+1}^n \dfrac{1}{k^2 + j^2}\\ &=s_1(n)+s_2(n)\\[2ex] s_1(n) &=\sum_{k=1}^n \sum_{j=1}^k \dfrac{1}{k^2 + j^2}\\ &=\sum_{k=1}^n \dfrac1{k^2}\sum_{j=1}^k \dfrac{1}{1 + (j/k)^2}\\ &=\sum_{k=1}^n \dfrac1{k}\dfrac1{k}\sum_{j=1}^k \dfrac{1}{1 + (j/k)^2}\\ &\approx\sum_{k=1}^n \dfrac1{k}\int_0^1\dfrac{dx}{1+x^2} \qquad\text{(Riemann sum)}\\ &=\sum_{k=1}^n \dfrac1{k}\dfrac{\pi}{4}\\ &=\dfrac{\pi}{4}\sum_{k=1}^n \dfrac1{k}\\ &\gt\dfrac{\pi}{4}\ln(n) \qquad\text{(Harmonic sum)}\\[2ex] s_2(n) &=\sum_{k=1}^n \sum_{j=k+1}^n \dfrac{1}{k^2 + j^2}\\ &\ge\sum_{k=1}^n \sum_{j=k+1}^n \dfrac{1}{k^2 + n^2}\\ &=\sum_{k=1}^n \dfrac{n-k}{k^2 + n^2}\\ &=\dfrac1{n}\sum_{k=1}^n \dfrac{1-k/n}{1+(k/n)^2}\\ &\approx \int_0^1 \dfrac{1-x}{1+x^2}dx\\ &=\dfrac14(\pi - \log(4))\\ &≈0.43882\\ \end{align}

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It is certainly true that there are some amazing identities (due to Euler and others) that reduce this question to simpler ones, there are also more direct, "geometric" ideas, perhaps roughly attributable to Minkowski in our times, though certainly people understood them earlier.

So, first, one might imagine some sort of integral test to apply here. In fact, with some trouble this can be justified, and it gives the "obvious" answer.

Somewhat more subtly-and-precisely, and which would amount to a proof of some version of an integral test in this case, on geometric grounds we can argue that any annulus of inner radius $r$ and outer radius $R$ contains at most (a constant times!?!) $\pi(R^2-r^2)$ (the area) lattice points. Then the conclusion is easy. Making this (or something equivalent) precise is not so hard.

The more difficult cases, due to Minkowski (at latest) involve error terms...

This approach applies as well to the $n$-dimensional case, to see that $\sum_{v\in \mathbb Z^n} 1/(v_1^2+\ldots+v_n^2)^s$ is absolutely convergent exactly for $\Re(s)>n/2$...

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Take the sum for n < m and for n >= m which is the same as the sum for n <= m because of symmetry.

We have $m^2 \le n^2 + m^2 \le 2m^2$, and $(m^2)^{-r/2} \ge (n^2 + m^2) ^{-r/2} \ge (2m^2 )^{-r/2} $. For every m there are 2m -1 values n < m or n <= m, so for every fixed m we have a sum between $2m (m^2) ^{-r/2}$ and $2m (2m^2) ^{-r/2}$ or between $2 m^{1-r}$ and $2^{1-r/2} m^{1-r}$.

The sum over all m converges iff the exponent 1-r is less than -1 or r> 2.

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